Answer to Question #259113 in Physical Chemistry for ceejay

Question

When baking soda (NaHCO3) is heated, carbon dioxide (CO2) is released. This 

makes bread, cookies and other pastries to rise. If 42.0 g of NaHCO3 was 

used, how much CO2 was released? (NaHCO3 = 84.00 g/mol, CO2 = 44.01, 

Na2CO3 = 105.99 g/mol, H2O = 18.02 g/mol)

2NaHCO3  Na2CO3 + H2O + CO2

Answer 1

The important thing to do here is write a balanced chemical equation for this decomposition reaction.

Sodium bicarbonate,

NaHCO3

, will decompose to form sodium carbonate,

Na2CO3

, water, and carbon dioxide

$CO 2 2 NaHCO 3(s] → Na 2 CO 3(s] + CO 2(g] + H 2 O$

Notice that you have a 2:1 mole ratio between sodium bicarbonate and sodium carbonate. This means that the reaction will produce half as many moles of the latter than whatever number of moles of the former underwent decomposition.

Use sodium carbonate's molar amss to determine how many moles you'd get in that sample

$0.685 g ⋅ 1 mole NaHCO 3 84.007 g = 0.008154 moles NaHCO 3$

Now, if the reaction were to have a100% yield, it would produce 0.008154 moles NaHCO3⋅1 mole Na2CO3

2 moles NaHCO3=0.004077 moles Na2CO3

Use the molar mass of sodium carbonate to determine how many grams would contain this many moles

0.004077moles⋅105.99 g

1mole=0.4321 g Na2CO3